Capacitor `C_(3)` in the circuit is a veriable capacitor (its capacitance can be varied). Group is plotted between potential difference `V_(1)` (across capacitor `C_(1)`) versus `C_(3)`. Electric potential `V_(1)` approaches to 10 V as `C_(3) rarr oo`. Given that `C_(1) + C_(2) = 10 mu F`

EMF of the battery is equal to

Options

A. 10V
B. 12V
C. 16V
D. 20V

Answer

B

Explanation

When `C_(3)toinfty` it means distance between the plates of `C_(3)` is zero or both plates will be at the same potential then `C_(2)` will be shorted entire pontential V will ber across `C_(1)` hence `V=V_(1)=10V`
From graphs when `C_(3)=0,V_(1)=2V. So C_(3)` will act like open switch `C_(1)` and `C_(2)` will be in series potential different across `C_(2)` is `V_(2)=10-V_(1)`
`=8V`
`q=C_(1)V_(1)=C_(2)V_(2)` or `C_(12)=C_(28)` or `C_(1)=4C_(2)`
`V_(1)=4V,V_(2)=10-4=6V`
`q=C_(1)V_(1)=(C_(2)+C_(3))V_(2)`
or `C_(14)=(C_(2)+C_(3))6` or `16C_(2)=6C_(2)+6C_(3)`
or `C_(3)=5C_(2//3)`